Supplementary Document for the Manuscript entitled A Functional Single Index Model

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1 Supplementary Document for the Manuscript entitled A Functional Single Index Model S1 Definition and Regularity Conditions Let H e the separale Hilert space of square integrale functions on, 1], and let < φ 1, φ 2 > φ 1t)φ 2 t)dt denote the inner product of the functions φ 1, φ 2 in H. Then the L 2 norm 2 is the norm induced y the inner product. Let α ) e a function in H. Let γ e the B-spline coefficient such that B r ) T γ converges to α ), uniformly on, 1] when the numer of the B-spline inner knots goes to infinite, and B r ) T γ converges to α ), the true function. Here B r = B r1,..., B rdγ ) T is the rth order B-spline asis vector. Note that for identifiaility, we assumed α ) = B r ) T γ = 1, i.e., γ 1 = B r1 ) 1 1 dγ k=2 B rk)γ k } = 1. So we only have d γ 1 B-spline coefficients to estimate, and we denote γ = γ 2,..., γ dγ ) T. Further, we denote αx) = Xt)αt)dt and α X) = Xt)α t)dt. We denote the functional sup norm as, and the L p norm as p. In addition, we define Γβ) to e a second moment ased linear operator such that Γβ)φt) Eβ T Xt)β T Xs)}φs)ds = E< β T X, φ > β T Xt)} while its empirical version can e written as Γ n β)φt) n 1 < β T X i, φ > β T X i t). Using these definitions, we can write < Γβ)B rk, B rl > = E B rk t)β T X i t)dt ) B rl t)β T X i t)dt,

2 < Γ n β)b rk, B rl > = n 1 B rk t)β T X i t)dt B rl t)β T X i t)dt. Define Cβ) as a d γ d γ matrix with its k, l) element < Γβ)B rk, B rl >, and Ĉβ) as a d γ d γ matrix matrix with its k, l) element < Γ n β)b rk, B rl >. The locally efficient score function for β, γ can e written as S effβ Y, Z, β, γ, g ) = g Y, β T Zγ) Eg Y, β T Zγ) β T Zγ}]Θ β Z EZ β T Zγ)}γ, S effγ Y, Z, β, γ, g ) = g Y, β T Zγ) Eg Y, β T Zγ) β T Zγ}]Θ γ Z EZ β T Zγ)} T β. Here g Y, β T Zγ) = f 2 Y, β T Zγ)/f Y, β T Zγ), f Y, β T Zγ) is the possily misspecified conditional density of Y given β T Zγ and f 2 Y, β T Zγ) is the derivative of f Y, β T Zγ) with respect to β T Zγ. When f is correctly specified, i.e. when f = f, we denote the resulting g as g. Θ β = I J 1, ) and Θ γ =, I dγ 1). Note that when f Y, β T Zγ) = fy, β T Zγ), the true density function, the locally efficient score is the efficient score of 2). Throughout the text, A 1 = o p A 2 ) for aritrary vectors or matrices A 1, A 2 means that A 1 has smaller order than A 2 component wise. We first list the regularity conditions under which we perform our theoretical analysis. A1) The kernel function K ) is non-negative, has compact support, and satisfies Ks)ds = 1, Ks)sds = and Ks)s 2 ds <, and K 2 s)ds <. A2) The andwidth h in the kernel smoothing satisfies nh 2 and nh 8 when n. A3) Assume α α C q, 1]), α is one-to-one, and α ) = 1}. The spline order r q. A4) We define the knots t r+1... t = and t N+r... t N+1 = 1. Let N e the numer of interior knots and, 1] e divided into N + 1 2

3 suintervals. In this case, d γ = N + r. Let h p e the distance etween the p + 1)th and pth interior knots of the order r B-spline functions and let h = that max r p N+r h p. There exists a constant C hn, < c h max h p = ON 1 ) and h / min h p < c h, r p N+r r p N+r <, such where N is the numer of knots which satisfies N as n, and N 1 nlogn) 1 and Nn 1/2q+1). A5) γ is a spline coefficient such that sup t,1] B r t) T γ α t) = O p h q ). The existence of such γ has een shown in De Boor 1978). A6) X j ), j = 1,..., J are continuous random functions in t, 1]. For each A7) t, EX j t)} <. β T X ) 2 is finite almost surely for any ounded β. Eβ T X i ) β T Z i γ} and its first two derivatives have finite L 2 norm almost surely for any ounded β, γ. The second moment operator Γβ) is strictly positive definite. ES effβ Y i, Z i, β, γ, g ) T, S effγ Y i, Z i, β, γ, g ) T } T ] is a smooth function of β T, γ T ) T and has unique root for β, γ. A8) f Y, β T Zγ) is a continuous density function and has ounded derivative with respect to the second argument. f Y, β T Zγ) is ounded away from and on its support. This implies g Y, β T Zγ) is a continuous ounded function and g Y, β T Zγ) is ounded away from and on its support. Let f Z β T Z i γ ) e the density for β T Zγ. f Z has two ounded derivatives, and f Z is ounded away from and on its support. 3

4 S2 Proofs of the propositions in the main text S2.1 Proof of Proposition 1 Assume the prolem is not identifiale. f, β, αt)} such that J } 1 f Y, β j αt)x j t)dt = f Y, Then there exist f, β, αt)} J } 1 β j αt)x j t)dt S1) for all Y and all functions X 1 ),..., X J ). Because S1) holds for all X j ), j = 1,..., J, it holds when X j t) = for j = 1,..., J 1 as well. Note that β J = β J = 1, we then have } f Y, αt)x J t)dt = f Let X J t) = at) + δt) and let δ, this yields f 2 = lim δ = lim δ = f 2 Y, f Y, f Y, Y, } αt)at)dt αt)t)dt ] αt)at) + δt)}dt f } Y, αt)x J t)dt. S2) Y, ] αt)at) + δt)}dt f Y, } αt)at)dt αt)t)dt, }) αt)at)dt /δ }) αt)at)dt /δ S3) where f 2Y, ), f 2Y, ) denote the partial derivatives of f, f with respect to the second component. Set t) to e the delta function with mass at. Because α) = α) = 1, this yields f 2 Y, } } αt)at)dt = f 2 Y, αt)at)dt for any a ), hence susequently from S3), αt)t)dt = αt)t)dt 4

5 for any ). Because αt), αt) are continuous, the latter directly yields αt) = αt). From S2), this further indicates fy, ) = fy, ). It is now easy to see from S1) that y taking X 1 t),..., X j 1 t), X j+1 t),..., X J t) to e zero, we can otain β j = β j for j = 1,..., J 1. This contradicts the assumption f, β, αt)} f, β, αt)}. Hence the model in 1) is indeed identifiale. S2.2 Proof of Proposition 2 Recall that we can write β T Z i γ as B rt) T γβ T X i t)dt. Note that ES effβ Y i, X i ); β, α ), g } T, S effγ Y i, X i ); β, α ), g } T ] T ) =, where S effβ Y i, X i ); β, α ), g } T, S effγ Y i, X i ); β, α ), g } T ] T is the vector which replaces the occurrence of B r ) T γ in the vector S effβ Y i, Z i, β, γ, g ) T, S effγ Y i, Z i, β, γ, g ) T ] T y α ). So the uniform convergence of B r ) T γ to α ) leads to ES effβ Y i, Z i, β, γ, g ) T, S effγ Y i, Z i, β, γ, g ) T } T = o1). On the other hand, from the estimating equation n 1 ŜeffβY i, Z i, β, γ, g ) T, ŜeffγY i, Z i, β, γ, g ) T } T =, using the consistency of the nonparametric kernel estimation, we have n 1 S effβ Y i, Z i, β, γ, g ) T, S effγ Y i, Z i, β, γ, g ) T } T = o p 1). The law of large numers further leads to ES effβ Y i, Z i, β, γ, g ) T, S4) S effγ Y i, Z i, β, γ, g ) T } T β= β,γ= γ = o p 1). The unique root property in Condition A7) implies that the derivative of ES effβ Y i, Z i, β, γ, g ) T, Y i, Z i, β, γ, g ) T }] T 5

6 with respect to β T, γ T ) T is a nonsingular matrix in the neighorhood where the function value is. Therefore, the left hand side of oth S4) and S4) are invertile functions of β T, γ T ) T. This implies β β = o p 1) and γ γ = o p 1) y the continuous mapping theorem. S2.3 Proof of Proposition 3 Denote the nuisance tangent space corresponding to f and the marginal density of Z as respectively Λ f and Λ z. Taking derivative of the loglikelihood function with respect to the nuisance parameters in each parametric sumodel and then taking the closure of their union, we have Λ z = fz) : f such that Ef) = } Λ f = fy, β T Zγ) : f such that Ef Z) = Ef β T Zγ) = }. We can easily verify that Λ z Λ f, hence Λ = Λ z Λ f, where Λ is the nuisance tangent space. Thus, Λ = Λ z Λ f. It is easy to see that Λ z = fy, Z) : Ef Z) = }. We now show that Λ f show this in two aspects. = fy, Z) : Ef Y, β T Zγ) = Ef β T Zγ)}. We First, it is easy to verify that functions having the aove conditional expectation property are elements in Λ f. To show the second aspect that elements in Λ f have to satisfy the conditional expectation requirement, consider any fy, Z) Λ f. We choose g = Ef Y, βt Zγ) Ef β T Zγ). Oviously, g Λ f hence Eg T f) =. We write this relation alternatively as = Eg T f) = Eg T Ef Y, β T Zγ)} = Eg T g) + Eg T Ef β T Zγ)} = Eg T g) + EEg β T Zγ) T Ef β T Zγ)} = Eg T g) +. This implies g itself should e zero. This means f indeed satisfies the conditional expectation requirement. 6

7 We are now ready to prove the form of Λ. For convenience, we denote set descried in the proposition A first and we will estalish Λ = A through proving Λ A and A Λ. To see A Λ, we can verify that for any fy, Z) Ef Y, β T Zγ) A, we oviously have EfY, Z) Ef Y, β T Zγ) Z} = Ef Z) Ef β T Zγ) =, thus A Λ z. The form of elements in A also ensures that EfY, Z) Ef Y, β T Zγ) Y, β T Zγ} = = EfY, Z) Ef Y, β T Zγ) β T Zγ}. Thus A Λ f. This shows A Λ. We now show Λ A. Assume gy, Z) Λ. Then Eg Z) = and Eg Y, β T Zγ) = Eg β T Zγ) = EEg Z) β T Zγ} =. This means we can always write gy, Z) as fy, Z) Ef Y, β T Zγ). The additional requirement of Eg Z) = further imposes Ef Z) = EEf Y, β T Zγ) Z} = Ef β T Zγ). Thus, we indeed have Λ A. S3 Proofs of the theorems in the main text S3.1 Proof of Theorem 1 By the consistency shown in Proposition 2, we expand the score function as = n 1/2 g Y i, β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )}g Y j, β T } Z j γβ )} K hβ T Z j γβ ) β T Z i γβ )} Θ γ Z i K } hβ T Z j γβ ) β T T Z i γβ )}Z j K β hβ T Z j γβ ) β T Z i γβ )} S5) = R + Tn 1/2 γ β ) γ }, S6) where J R = n 1/2 g Y i, β T Z i γ ) K } hβ T Z j γ β T Z i γ )g Y j, β T Z j γ ) K hβ T Z j γ β T Z i γ ) Θ γ Z i K } hβ T Z j γ β T T Z i γ )Z j K β hβ T Z j γ β T, Z i γ ) 7

8 T = n 1 g Y i, β T Z i γ) K } hβ T Z j γ β T Z i γ)g Y j, β T Z j γ) K hβ T Z j γ β T Z i γ) Θ γ Z i K } hβ T Z j γ β T T ] Z i γ)z j K β hβ T Z j γ β T / β T Z i γ) Z i γ) γ=γ β T Z i Θ T γ, and γ is the point on the line connecting γβ ) and γ. The Asymptotic Property of R Now R can e written as R = R + R 1 + R 2 + R 3, where R = n 1/2 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ Z i EZ i β T Z i γ ) } T β, R 1 = n 1/2 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZ i β T Z i γ ) K } hβ T Z j γ β T T Z i γ }Z j K β hβ T Z j γ β T, Z i γ } R 2 = n 1/2 Eg Y i, β T Z i γ ) β T Z i γ } K hβ T Z j γ β T Z i γ )g Y j, β T Z j γ ) K hβ T Z j γ β T Z i γ ) Z i EZ i β T Z i γ ) } T β, R 3 = n 1/2 Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ K hβ T Z j γ β T Z i γ )g Y j, β T ] Z j γ ) K Θ hβ T Z j γ β T γ Z i γ ) EZ i β T Z i γ ) K } hβ T Z j γ β T T Z i γ }Z j K β hβ T Z j γ β T. Z i γ } Further, we write R 1 = R 11 + R 12, where R 11 = n 1/2 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ ) K hβ T Z j γ β T Z i γ ) nf Z β T Z i γ ) 8

9 K hβ T Z j γ β T } Z i γ }Z T j β nf Z β T, Z i γ ) R 12 = n 1/2 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ ) K hβ T Z j γ β T Z i γ } nf Z β T Z i γ ) K hβ T Z j γ β T Z i γ )Z j } T β nf Z β T Z i γ ) } nf Z β T Z i γ ) K hβ T Z j γ β T Z i γ ) 1 } nf Z β T Z i γ = R ) 11 K hβ T Z j γ β T Z i γ ) 1, where f Z β T Z i γ ) is the density for β T Zγ. We first analyze R 11. Using the U statistics property, we write R 11 as R 11 = R R 112 R o p R R 112 R 113 ), where where R 111 = R 11 = n 3/2 J g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T } T Z i γ )Z j β nf Z β T Z i γ ) = R R 112 R o p R R 112 R 113 ), n 1/2 g E Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T } T ) Z i γ )Z j β nf Z β T O i, Z i γ ) J g R 112 = n 1/2 E Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ 9

10 EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T } T ) Z i γ )Z j β O j, nf Z β T Z i γ ) g R 113 = n 1/2 E Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T Z i γ )Z j nf Z β T Z i γ ) } T β )}, where O i is the random variale corresponding to the ith oservation. Now ecause the summand for R 111, i.e., g E Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T } T ) Z i γ )Z j β nf Z β T O i Z i γ ) = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )EK h β T Z j γ β T Z i γ ) O i } f Z β T Z i γ ) EK hβ T Z j γ ) β T ] T Z i γ )Z j O i } β f Z β T Z i γ ) = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )EK h β T Z j γ β T Z i γ ) O i } f Z β T Z i γ ) EK hβ T Z j γ β T Z i γ )EZ j β T ] T Z j γ ) O i ] β f Z β T Z i γ ) = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZ i β T Z i γ ) T Ks)s 2 β 1 + 2f Z β T Z i γ ) 2 f Z aβ T } Z i γ, sh)} dsh 2 EZ aβ T i β T Z i γ, sh) 2 Z i γ ) T β 2 EZ i β T β T Z i γ, sh)} T β f Z β T Z i γ, sh)}] β T Z i γ, sh) 2 dsh 2 1 s 2 Ks) 2f Z β T Z i γ ) ]

11 = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZ i β T Z i γ ) T 2 f Z aβ T Z i γ β, sh)} s 2 Ks) aβ T Z i γ, sh) 2 2f Z β T Z i γ ) ds 2 EZ i β T Z i γ, sh)} T β f Z β T ] Z i γ, sh)}] s 2 Ks) β T Z i γ, sh) 2 2f Z β T Z i γ ) ds h 2, S7) where aβ T Z i γ, sh), β T Z i γ, sh) are points on the line connecting β T Z i γ and β T Z i γ +sh. Note that aβ T Z i γ, sh), β T Z i γ, sh) depend on the values of oth β T Z i γ and sh and will go to β T Z i γ when h. Now note that g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZ i β T Z i γ ) T 2 f Z aβ T Z i γ β, sh)} s 2 Ks) aβ T Z i γ, sh) 2 2f Z β T Z i γ ) ds 2 EZ i β T Z i γ, sh)} T β f Z β T Z i γ, sh)}] β T Z i γ, sh) 2 can e written as with C 1i t) C 1i t)θ γ B r t)dt = C 1i t)b rk t), k = 2,..., d γ } T dt, s 2 Ks) 2f Z β T Z i γ ) ds = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] EX i t) T β β T 2 f Z aβ T Z i γ Z i γ ), sh)} s 2 Ks) aβ T Z i γ, sh) 2 2f Z β T Z i γ ) ds 2 EX i t) T β β T Z i γ, sh)}f Z β T ] Z i γ, sh)}] s 2 Ks) β T Z i γ, sh) 2 2f Z β T Z i γ ) ds. Oviously EC 1i t)} = and hence EC 1i t)} <. From Conditions A6) and A8), C 1i ) 2 <, a.s., so y Lemma 3, we have ] R 111 = O p h 2 h logn)}, S8) which also implies R 113 = ER 111 ) = Oh 2 h logn)}. S9) 11

12 Further, the summand of R 112, i.e., g E Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ γ EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T } T ) Z i γ )Z j β nf Z β T O j Z i γ ) = E E g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] ) β T Z i γ, O j Θγ EZi β T Z i γ )K h β T Z j γ β T Z i γ ) nf Z β T Z i γ ) K hβ T Z j γ β T } T ) Z i γ )Z j β nf Z β T O j Z i γ ) =. Comining S8), S1), S9), we get S1) R 11 = O p h 2 h logn)}. S11) Further, to treat R 12, note that } nf Z β T Z i γ ) K hβ T Z j γ β T Z i γ ) 1 hence = fz β T Z i γ ) n 1 J K } hβ T Z j γ β T Z i γ ) n 1 K hβ T Z j γ β T Z i γ ) = O p h 2 + nh) 1/2 } S12) R 12 = o p h 2 h logn)}. S13) Thus, comining S11) and S13), we have R 1 = O p h 2 h logn)}. S14) 12

13 Using similar derivation as those led to S14) while exchange the roles of Z and g, we can otain R 2 = O p h 2 h logn)}. S15) For R 3, we first note that Eg Y i, β T Z i γ ) β T Z i γ } K ] hβ T Z j γ β T Z i γ )g Y j, β T Z j γ ) K hβ T Z j γ β T Z i γ ) = O p h 2 + n 1/2 h 1/2 ) y the uniform consistency of the kernel estimator. Further, the summand in R 3 has the form Eg Y i, β T Z i γ ) β T Z i γ } K ] hβ T Z j γ β T Z i γ )g Y j, β T Z j γ ) K hβ T Z j γ β T Z i γ ) EZ i β T Z i γ ) K } hβ T Z j γ β T T Z i γ }Z j K β hβ T Z j γ β T Z i γ } = = where h 4 + n 1 h 1 )C 2i t)θ γ B r t)dt h 4 + n 1 h 1 )C 2i t)b rk t), k = 2,..., d γ } T dt, C 2i t) = h 4 + n 1 h 1 ) 1 Eg Y i, β T Z i γ ) β T Z i γ } K hβ T Z j γ β T Z i γ )g Y j, β T ] Z j γ ) K hβ T Z j γ β T Z i γ ) EX i t) T β β T Z i γ } K } hβ T Z j γ β T Z i γ }X j t) T β K. hβ T Z j γ β T Z i γ } Note that EC 2i t)}, EC 2i t)} <, C 2i ) 2 <, a.s. y Conditions A2), A6) and the uniform consistency of the kernel estimator. So y Lemma 3, we have R 3 = n 1/2 h 4 + n 1 h 1 )C 2i B rk t), k = 2,..., d γ } T dt 13 Θ γ

14 = O p n 1/2 h 4 + n 1/2 h 1 )h }. S16) We now assess S7). We write where R = n 1/2 g Y i, β T Z i γ } Eg Y i, β T Z i γ ) β T Z i γ }]Θ γ Z i EZ i β T Z i γ ) } T β = n 1/2 C 3i t)θ γ B r t)dt = n 1/2 C 3i t)b rk t), k = 2,..., d γ } T dt, C 3i t) = g Y i, β T Z i γ } Eg Y i, β T Z i γ ) β T Z i γ }]X i t) T β EX i t) T β β T Z i γ }]. We can easily verify that EC 3i t)} =, EC 3i t)} < and C 3i ) 2 <, a.s. y Condition A6). Therefore, y Lemma 3 we know R = O p n 1/2 h ). S17) Comparing S14), S15), S16) with S17), from Condition A2), it is clear that R dominates R 1, R 2 and R 3. We further analyze R y writing where R = R + R 1 + R 2 S18) R = n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ and Z i EZ i β T α X i )} ] T β S19) R 1 = n 1/2 g Y i, β T Z i γ } Eg Y i, β T Z i γ } β T Z i γ }] 14

15 ) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] Θ γ Z i EZ i β T Z i γ } ] T β, R 2 = n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T Z i γ } ] Z i EZ i β T α X i )} ] ) Tβ. To analyze R, we write it as R = n 1/2 = n 1/2 C 4i t)θ γ B r t)dt C 4i t)b rk t), k = 2,..., d γ } T dt, where C 4i t) = g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]X i t) T β EX i t) T β β T α X i )}]. We can check that EC 4i t)} = and C 4i ) 2 < a.s. y Condition A6). Thus, y Lemma 3, R = h logn). S2) Further R 2 = n 1/2 = n 1/2 h q C 5it)Θ γ B r t)dt h q C 5it)B rk t), k = 2,..., d γ } T dt, where C 5i t) = h q g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] 15

16 X i t) T β EX i t) T β β T Z i γ }] X i t) T β ) EX i t) T β β T α X i )}]. Now EC 5i t)} = ecause E g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] Xi t) T β EX i t) T β β T Z i γ } ] X i t) T β )} EX i t) T β β T α X i )}] } = E E g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] X i Xi t) T β EX i t) T β β T Z i γ } ] X i t) T β )] EX i t) T β β T α X i )}] =. Further C 5i ) 2 <, a.s. y Condition A6). Therefore, y Lemma 3 we have R 2 = n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T Z i γ } ] Z i EZ i β T α X i )} ] ) Tβ = O p h q h logn)}. In addition, where R 1 = n 1/2 C 6i t) = h q = n 1/2 h q C 6it)Θ γ B r t)dt h q C 6it)B rk t), k = 2,..., d γ } T dt, g Y i, β T Z i γ } Eg Y i, β T Z i γ } β T Z i γ }] 16

17 ) g Y i, β T α X i )} Eg Y i, β T Z i γ } β T α X i )}] X i t) T β EX i t) T β β T Z i γ } ], with EC 6i t)}, EC 6i t)} <, C 6i ) 2 <, a.s. y Condition A6). Therefore, y Lemma 3 we have R 1 = n 1/2 g Y i, β T Z i γ } Eg Y i, β T Z i γ } β T Z i γ }] ) g Y i, β T α X i )} Eg Y i, β T Z i γ } β T α X i )}] Θ γ Z i EZ i β T Z i γ } ] T β = O p h q+1 n 1/2 ). Now y the fact that R 1 + R 2 = O p h q+1 n 1/2 + h q h logn)} and R = O p h logn)} y S2), we have R 1 + R 2 = o p R ) y Condition A4). Therefore, y S18) we have R = R +o p R ), comining with the fact that R dominates R 1, R 2, R 3 and R = R + R 1 + R 2 + R 3 in S5), we have R = R + o p R ), S21) i.e., n 1/2 J g Y i, β T Z i γ } K } hβ T Z j γ β T Z i γ }g Y j, β T Z j γ } K hβ T Z j γ β T Z i γ } Θ γ Z i K } hβ T Z j γ β T T Z i γ }Z j K β hβ T Z j γ β T = R + o p R ), Z i γ } where R is given in S19). The Asymptotic Property of T T = n 1 g Y i, β T Z i γ) Now consider the term K hβ T Z j γ β T Z i γ)g Y j, β T } Z j γ) K hβ T Z j γ β T Z i γ) 17

18 Θ γ Z i K } hβ T Z j γ β T T ] Z i γ)z j K β hβ T Z j γ β T Z i γ) ) / β T Z i γ) γ=γ β T Z i Θ T γ S22) in S5). We decompose T as T = T + T 1 + T 2, where T = T 1 = T 2 = n 1 g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] Θ γ Z i EZ i β T α X i )} ] } T β )/ β T α X i )}β T Z i Θ T γ, g n 1 Y i, β T Z i γ) Eg Y i, β T Z i γ) β T Z i γ} ] Θ γ Z i EZ i β T Z i γ} ] } T β )/ β T Z i γ) γ=γ β T Z i Θ T γ g n 1 Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] Θ γ Z i EZ i β T α X i )} ] } T β )/ β T α X i )}β T Z i Θ T γ, n 1 g Y i, β T Z i γ) K } hβ T Z j γ β T Z i γ)g Y j, β T Z j γ) K hβ T Z j γ β T Z i γ) Θ γ Z i K } hβ T Z j γ β T T ] ) Z i γ)z j K β hβ T Z j γ β T / β T Z i γ) Z i γ) γ=γ β T Z i Θ T γ g n 1 Y i, β T Z i γ) Eg Y i, β T Z i γ) β T Z i γ} ] Θ γ Z i EZ i β T Z i γ} ] } T β )/ β T Z i γ) γ=γ β T Z i Θ T γ. Now T = max k=1,...,d γ d γ n 1 l=1 C 7i t)b rk t)}dt 18

19 = max k=1,...,d γ β T X i s)b rl s) B rl )/B r1 )B r1 s)}ds] d γ n 1 l=1 C 7i t)b rk t)}dt β T X i ξ 1 )B rl ξ 1 ) B rl )/B r1 )B r1 ξ 1 )}] = max k=1,...,d γ n 1 C 7i t)b rk t)}dt ] d γ β T X i ξ 1 ) B rlξ 1 ) B rl )/B r1 )B r1 ξ 1 )} = max k=1,...,d γ n 1 l=1 β T X i ξ 1 ) M = max k=1,...,d γ n 1 where ξ 1 is a point in, 1), and M = d γ l=1 d γ l=1 <, C 7i t)b rk t)}dt a n C 7i t)b rk t)}dt, B rlξ 1 ) B rl )/B r1 )B r1 ξ 1 )} B rl ξ 1 ) + d γ l=1 B rl )/B r1 )B r1 ξ 1 ) which holds y 3.4) on page 141 in DeVore and Lorentz 1993), and a n is the sequence such that β T X i ξ 1 ) = O a.s. a n ). Note that a n does not need to e ounded. Here C 7i t) = g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] X i t) T β EX i t) T β β T α X i )} ] T β )/ β T α X i )} and g C 7i t) = a 1 n Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] 19

20 X i t) T β EX i t) T β β T α X i )} ] T β )/ β T α X i )} Mβ T X i ξ 1 ). Now EC 7i t)} =, EC 7i t)} <, and C 7i ) 2 <. By Lemma 3, we have T = O p h a n ). Similarly, T 1 = max k=1,...,d γ n 1 h q a nc 8i t)b rk t)}dt where g C 8i t) = h q a 1 n Y i, β T Z i γ) Eg Y i, β T Z i γ) β T Z i γ} ] X i t) T β EX i t) T β β T Z i γ} ] ) / β T Z i γ) γ=γ Mβ T X i ξ 1 ) g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] X i t) T β EX i t) T β β T α X i )} ] ) } / β T α X i )}Mβ T X i ξ 1 ), with EC 8i t)} =, EC 8i t)} <, and C 8i ) 2 <. By Lemma 3, we have T 1 = O p h q+1 a n ). Next, T 2 = max k=1,...,d γ n 1 a n h 2 + n 1/2 h 1/2 )C 9i t)b rk t)}dt, where C 9i t) = a 1 n h 2 + n 1/2 h 1/2 ) 1 g Y i, β T Z i γ) K hβ T Z j γ β T Z i γ)g Y j, β T } Z j γ) K hβ T Z j γ β T Z i γ) X i t) T β K } ] hβ T Z j γ β T Z i γ)x j t) T β K / β T hβ T Z j γ β T Z i γ) Z i γ) g Mβ T X i ξ 1 ) Y i, β T Z i γ) Eg Y i, β T Z i γ) β T Z i γ} ] X i t) T β EX i t) T β β T Z i γ} ] ) ) / β T Z i γ) γ=γ Mβ T X i ξ 1 ), γ=γ 2

21 with EC 9i t)} =, EC 9i t)} <, and C 9i ) 2 <. By Lemma 3, we have T 2 = O p h h 2 +n 1/2 h 1/2 )a n }. Comining the orders of T, T 1 and T 2, we see that T clearly dominates T 1, T 2. So we can write T = T +o p T ). We further analyze T. Note that T = T + T 1 + T 2, where T = = T 1 = n 1 g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] Z i EZ i β T α X i )} ] T β β T n 1 β T α X i ) Θ γ Zi EZ i β T α X i )} ] T β Zi EZ i β T α X i )} ] Θ T γ g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] β T α X i ) Zi EZ i β T α X i )} ] ) Θ T γ, n 1 g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] β T ) Θ γ / β T α X i )}Θ γ Zi EZ i β T α X i )} ] T β )β T } EZ i β T α X i )}]Θ T γ, and T 2 = n 1 g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] Θ γ Zi EZ i β T α X i )} ] ) } T β / β T α X i )}β T Z i Θ T γ. Now where T = max k=1,...,d γ n 1 a n C 1i t)b rk t)}dt, C 1i t) = a 1 n Xi t) T β EX i t) T β β T α X i )} ] 21

22 g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] M β T α X i ) β T X i ξ 1 ) EX i ξ 1 ) β T α X i )}]. Now EC 1i t)}, EC 1i t)} < and C 1i ) 2 <. By Lemma 3, we have T = O p h a n ). Similarly T 1 + T 2 = max k=1,...,d γ n 1 a n C 11i t)b rk t)}dt, where g C 11i t) = a 1 n Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] / β T α X i )}Θ γ Xi t) T β EX i t) T β β T α X i )} ] ) Mβ T EX i ξ 1 ) β T α X i )}] + a 1 n g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )}] X i t) T β ) EX i t) T β β T α X i )}] / β T α X i )}Mβ T X i ξ 1 ). Now EC 11i t)} = and C 11i ) 2 <. By Lemma 3, we have T 1 + T 2 = O p h n 1 logn)a n } = o p T ) S23) y Condition A4). Comining the results that T = T + o p T ) and T = T + o p T ), we have T = T + o p T ). S24) The Asymptotic Property of γβ ) S5), S21) and S24), we otain Comining the aove results with n 1/2 γ β ) γ } = T 1 R 1 = T + o p T )} R + o p R )} 22

23 = E Θ γ Zi EZ i β T α X i )} ] T β g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] β T α X i ) β T Zi EZ i β T α X i )} ] Θ T γ 1 T + o p T )} R + o p R )} +E Θ γ Zi EZ i β T α X i )} ] T β ) 1 R + o p R )} g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] β T α X i ) ) 1 R + o p R )} β T Zi EZ i β T α X i )} ] Θ T γ = Θ γ C Q β )Θ T γ } 1 n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Z i EZ i β T α X i )} } T β + o p R ) Θ γ Ĉ Q β )Θ T γ + o p Θ γ Ĉ Q β )Θ T γ }] 1 n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T α X i )} } T β ) +o p R ) + Θ γ C Q β )Θ Tγ } 1 n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] Θ γ Zi EZ i β T α X i )} } ) T β + o p R ) ) = Θ γ C Q β )Θ T γ } 1 + o p Θ γ C Q β )Θ T γ } 1 ] n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] Θ γ Zi EZ i β T α X i )} } ) T β + o p R ), S25) 23

24 where ĈQβ ), C Q β ) are defined in Corollary 1 for v = β and the corresponding Q i t) g Y i, β T α X i )) Eg Y i, β T α X i )) β T α X i )} ] β T α X i ) X i t) EX i t) β T α X i )}]. ) 1/2 The last equality holds ecause for aritrary d γ 1 -dimensional unit vector u u 1,..., u γ 1 ) T, we have u T Θ γ = γ 1 k=1 u kb rk+1 )/B r1 ), u 1,..., u γ 1 } T. Now since B rk+1 ) is positive for at most r k s, γ 1 k=1 u kb rk+1 )/B r1 ) is finite, which implies Θ T γ u 2 2 = 1 u γ 1 k=1 u kb rk+1 )/B r1 )} 2 and Θ γ 2 = O p 1). Therefore, D 7 Θ T γ u 2 h u T Θ γ Ĉ Q β )Θ T γ u D 8 Θ T γ u 2 h S26) for ounded positive constants D 7, D 8 y Corollary 1, which implies the maximum and minimum eigenvalue of Θ γ Ĉ Q β )Θ γ is ounded etween D 7 Θ T γ u 2 h and D 8 Θ T γ u 2 h. Since Θ γ Ĉ Q β )Θ γ is a symmetric matrix, Θ γ Ĉ Q β )Θ γ 2 is its maximum eigenvalue, which is of order O p h ). Now note that Θ γ Ĉ Q β )Θ γ } 1 2 is the inverse of the minimum eigenvalue of Θ γ Ĉ Q β )Θ γ. Therefore, Θ γ Ĉ Q β )Θ γ } 1 2 = O p h 1 ). In summary, Θ γ Ĉ Q β )Θ γ 2 = O p h ), Θ γ Ĉ Q β )Θ γ } 1 2 = O p h 1 ) S27) Similar to those arguments that led to S27), we have Θ γ C Q β )Θ γ 2 = O p h ), Θ γ C Q β )Θ γ } 1 2 = O p h 1 ). S28) Now ecause as shown in Corollary 1, ĈQβ ) C Q β ) 2 o p h n h 1)/2 } and Θ γ 2 = O p 1), we have Θ γ Ĉ Q β )Θ T γ + o p Θ γ Ĉ Q β )Θ T γ } Θ γ C Q β )Θ T γ 2 24

25 Therefore, Θ γ Ĉ Q β )Θ T γ Θ γ C Q β )Θ T γ 2 + o p Θ γ Ĉ Q β )Θ T γ 2 } o p h n h 1)/2 } + o p h ). Θ γ Ĉ Q β )Θ T γ + o p Θ γ Ĉ Q β )Θ T γ }] 1 Θ γ C Q β )Θ T γ } 1 2 = Θ γ C Q β )Θ T γ } 2 Θ γ Ĉ Q β )Θ T γ + o p Θ γ Ĉ Q β )Θ T γ } Θ γ C Q β )Θ T γ ]1 + o p 1)} 2 o p h 2 h n h 1)/2 } + o p h 2 h ) = o p h 1 n h 1)/2 } + o p h 1 ) = o p Θ γ C Q β )Θ T γ } 1 ]. S29) As a result, S25) holds and we can write n 1/2 γ β ) γ } = Θ γ C Q β )Θ T γ } 1 + o p Θ γ C Q β )Θ T γ } 1 ]) n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T α X i )} ] ) T β + o p R ) = E Θ γ Z i EZ i β α X i }] T β g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ] ) β T α X i ) )} 1 ) β T Z i EZ i β T α X i )}]Θ T γ + o p Θ γ C Q β )Θ T γ } 1 ] n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Z i EZ i β T α X i )} ] ) T β + o p R ) = L + o p L), where recall that we have defined in the statement that L = Θ γ C Q β )Θ T γ } 1 n 1/2 g Y i, β T α X i )} 25

26 Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T α X i )} ] ) T β = E Θ γ Z i EZ i β α X i }] T β S3) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ] ) β T α X i ) )} 1 ) β T Z i EZ i β T α X i )}]Θ T γ n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T α X i )} ] T β ). Now for aritrary d γ 1 -dimensional vector a with a 2 < y using the same argument as those lead to S23), we have where a T n 1/2 γ β ) γ }] 2 = a T LL T a + o p a T LL T a) = L 1 + o p L 1 ), a T Θ L 1 = a T Θ γ C Q β )Θ T γ } 1 2 γ C Q β )Θ T γ } 1 2 a T Θ γ C Q β )Θ T γ } 1 2 n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Z i EZ i β T α X i )}] T β ) 2 Θ γ C Q β )Θ T γ } 1 a a T Θ γ C Q β )Θ T γ } 1 2 a T Θ = a T Θ γ C Q β )Θ T γ } 1 2 γ C Q β )Θ T γ } 1 2 a T Θ γ C Q β )Θ T γ } 1 2 n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Z i EZ i β T α X i )}} T β ) 2 O p h 2 h ) = O p h 1 ). Θ γ C Q β )Θ T γ } 1 a a T Θ γ C Q β )Θ T γ }

27 The third line holds y noting that Θ γ C Q β )Θ T γ } 1 = O p h 1 ) y S28) and n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]Θ γ Z i EZ i β T α X i )} } T β ) 2 is in the form of Θ γ Ĉ Q β )Θ T γ, where ĈQβ ) is defined in Corollary 1 with v = β and Q i t) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] X i t) EX i t) β T α X i )}]. Using the arguments as those led to S26), we have a T Θ γ C Q β )Θ T γ } 1 D 1 h n 1 g Y a T Θ γ C Q β )Θ T γ } 1 i, β T α X i )} 2 Eg Y i, β T α X i )} β T α X i )}]Θ γ Zi EZ i β T α X i )} } T β ) 2 Θ γc Q β )Θ T γ } 1 a a T Θ γ C Q β )Θ T γ } 1 2 D 11 h, S31) where D 1 and D 11 are positive constants. Therefore a T γ β ) γ } = O p n 1/2 h 1/2 ). This proves the result. S3.2 Proof of Theorem 2 The proof of Theorem 2 is divided into two parts. In Part I, we show that n β β) convergences to a random quantity which exists under the approximated model. In Part II, we show that this quantity converges to a Gaussian vector defined on the measurale space under the true model. Part I By the consistency shown in Proposition 2, we expand the score function as = n 1/2 g Y i, β T Z i γ β)} 27

28 where K h β T Z j γ β) β T Z i γ β)}g Y j, β T ] Z j γ β)} K h β T Z j γ β) β Θ T β Z i γ β)} Z i K h β T Z j γ β) β T Z i γ β)}z j K h β T Z j γ β) β γ β) T Z i γ β)} = G + Hn 1/2 β β ), S32) G = n 1/2 g Y i, β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )}g Y j, β T ] Z j γβ )} K Θ hβ T Z j γβ ) β T β Z i γβ )} Z i K ] hβ T Z j γβ ) β T Z i γβ )}Z j K γβ hβ T Z j γβ ) β T ) Z i γβ )} and H = n 1 g Y i, β T Z i γβ)} K hβ T Z j γβ) β T Z i γβ)}g Y j, β T ] Z j γβ)} K Θ hβ T Z j γβ) β T β Z i γβ)} Z i K ] hβ T Z j γβ) β T Z i γβ)}z j K γβ)/ β T hβ T Z j γβ) β T β=β, Z i γβ)} where β is the point on the line connecting β and β. In the following, we study the asymptotic properties for G and H separately. The Asymptotic Property for G Now we can decompose G as G = G + G 1 + G 2 + G 3, where G = n 1/2 g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β Z i EZ i β T Z i γβ )} ] γβ ), 28

29 G 1 = n 1/2 g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZ i β T Z i γβ )} K ] hβ T Z j γβ ) β T Z i γβ )}Z j K γβ hβ T Z j γβ ) β T ), Z i γβ )} G 2 = n 1/2 Eg Y i, β T Z i γβ )} β T Z i γβ )] K hβ T Z j γβ ) β T Z i γβ )}g Y j, β T Z j γβ )} K hβ T Z j γβ ) β T Z i γβ )} Z i EZ i β T Z i γβ )} ] γβ ), G 3 = n 1/2 Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β K hβ T Z j γβ ) β T Z i γβ )}g Y j, β T ) Z j γβ )} K Θ hβ T Z j γβ ) β T β Z i γβ )} EZ i β T Z i γβ )} K ] hβ T Z j γβ ) β T Z i γβ )}Z j K γβ hβ T Z j γβ ) β T ). Z i γβ )} We can further write G 1 = G 11 + G 12, where G 11 = n 1/2 g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZi β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T } Z i γβ )}Z j γβ nf Z β T ), Z i γβ )} G 12 = n 1/2 g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZi β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T } Z i γβ )}Z j nf Z β T Z i γβ )} } nf Z β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )} 1 γβ ) } nf Z β T Z i γβ = G )} 11 K hβ T Z j γβ ) β T Z i γβ )} 1. 29

30 Note that G 11, G 12 are of finite dimension, so the sup norm and the L 2 norm are equivalent. Using the U-statistics techniques similar to those led to S11) and S13), we get G 11 = n 3/2 where J g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZi β T Z i γβ )}K h β T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T } Z i γβ )}Z j γβ nf Z β T ) Z i γβ )} = G G 112 G o p G G 112 G 113 ), G 111 = n 1/2 g E Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZi β T Z i γβ )}K h β T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T ] Z i γβ )}Z j } γβ nf Z β T ) O i, Z i γβ )} J g G 112 = n 1/2 E Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZi β T Z i γβ )}K h β T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T ] Z i γβ )}Z j } γβ nf Z β T ) O j, Z i γβ )} g G 113 = n 1/2 E Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZi β T Z i γβ )}K h β T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )}Z j nf Z β T Z i γβ )} } γβ ) Similar to R 111, using the same derivation in S7), we can write the summand of G 111 as g E Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β ]. 3

31 EZi β T Z i γβ )}K h β T Z j γβ ) β T Z i γβ )} nf Z β T Z i γβ )} K hβ T Z j γβ ) β T ] Z i γβ )}Z j } γβ nf Z β T ) O i Z i γβ )} = g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β EZ i β T 2 f Z cβ T Z i γβ Z i γβ )} γβ ) ), sh}] s 2 Ks) cβ T Z i γβ ), sh} 2 2f Z β T Z i γβ )} ds 2 EZ i dβ T Z i γβ ), sh}} γβ )f Z dβ T Z i γβ ), sh}}] dβ T Z i γβ ), sh} ) 2 s 2 Ks) 2f Z β T Z i γβ )} ds h 2, which is a mean finite dimensional vector. Here cβ T Z i γβ ), sh} and dβ T Z i γβ ), sh} are points on the line connecting β T Z i γβ ) and β T Z i γβ )+ sh. Therefore, we have h 2 G 111 = O p 1), and in turn G 111 and G 113 have the order of O p h 2 ). Further, similar to R 112, we have G 112 =. As a result, G 11 = G G 112 G o p G G 112 G 113 ) = O p h 2 ). Now similar to S12), since } nf Z β T Z i γβ )} K hβ T Z j γβ ) β T Z i γβ )} 1 = O p h 2 + nh) 1/2 }, we otain G 12 = o p G 11 ). Thus, we have G 1 2 = G 11 + G 12 2 = O p h 2 ). S33) Using similar derivation as those led to S33) while exchanging the roles of Z and g, we otain G 2 2 = O p h 2 ). Using the same reasoning as those led to S16), and noting that G 3 is a finite dimensional vector, we have that the order of G 3 is the same as n 1/2 times the square of the order for kernel estimation errors, i.e., G 3 2 = O p n 1/2 h 4 + n 1/2 h 1 ) = o p 1) under Condition A2). Now we can further write G = G + G 1 + G 2, where G = n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ) Θ β α X i ) Eα X i ) β T α X i )} ], 31

32 G 1 = n 1/2 g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β Z i EZ i β T Z i γβ )} ] γβ ) n 1/2 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ }]Θ β Zi EZ i β T Z i γ ) } γ, G 2 = n 1/2 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ β Z i EZ i β T Z i γ )}γ n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )])Θ β α X i ) Eα X i ) β T α X i )} ] = n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ) Θ β αx i ) EαX i ) β T αx i )} ] α=α αx i ) X i,z i )}Z i γ α X i )}, where α X i, Z i ) is the point on the line connecting α X i ) and Z i γ. Clearly we have EG 2 ) =. Further from B r ) T γ α ) = O p h q ), we have = Z i γ α X i ) 2 B r t) T γ X i t)dt α t)x i t)dt 2 1/2 B r t) T γ α t) dt} 2 X i ) 2 = O p h q ), so h q G 2 2 = O p 1). Therefore, G 2 2 = O p h q ). Next = G 1 n 1 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] Θ β Zi EZ i β T Z i γ ) } γ ) n 1/2 γ β ) γ } 1 + o p 1)} γ T = G T 11n 1/2 γ β ) γ }1 + o p 1)}, 32

33 where G T 11 = n 1 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] β T Z i EZ i β T Z i γ )}Θ T γ. β T Z i γ Θ β Zi EZ i β T Z i γ ) } γ The last equality holds y using the same arguments as those lead to S23) Now G T 11 = n 1 is a J 1) d γ 1) matrix where C 12i t)b T r t)θ T γ dt C 12i t) = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] β T X i t) Eβ T X i t) β T Z i γ }]. β T Z i γ Θ β Zi EZ i β T Z i γ ) } γ Because J is finite, G 11 and G 11l have the same order as their elements did, where G 11l is the lth column of G 11. The k, l) element of G 11 is G 11kl = n 1 B rk t)}c 12li dt, where C 12li is the lth element of C 12i. Now EC 12li t)}, EC 12li t)} <, C 12li ) 2 <, a.s. y Condition A6). Therefore, y Lemma 3 we have G 11kl = O p h ), and thus G 11l and G 11 satisfy G 11l = O p h ), G 11 = O p h ). S34) Since a 2 N 1/2 a for aritrary vector a with length N, and 2 N 1/2 for aritrary matrix with N rows, we otain G 11l 2 N 1/2 O p h ) = O p h 1/2 ), G 11 2 N 1/2 O p h ) = O p h 1/2 ). 33

34 Therefore, G T 11n 1/2 γ β ) γ } = O p max 1 l J GT 11ln 1/2 γ β ) γ }] = h 1/2 O p max 1 l J h 1/2 G T 11ln 1/2 γ β ) γ }] = O p 1) y Theorem 1 that a T γ β ) γ } = O p n 1/2 h 1/2 ) for aritrary a with a 2 <. Further, G T 11n 1/2 γ β ) γ } = G T 13n 1/2 γ β ) γ } + G 11 G 12 ) T n 1/2 γβ ) γ } where +G 12 G 13 ) T n 1/2 γ β ) γ } G T 12 g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] = E Θ β T β Zi EZ i β T Z i γ ) } γ Z i γ ) β T Z i EZ i β T Z i γ )}Θ T γ and G T 13 = g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ) E β T α X i )} Therefore, α X i ) Eα X i ) β T α X i )} ] β T Z i EZ i β T α X i )}]Θ T γ Θ β }. where G T 11 G T 12 = n 1 C 13i t)b T r t)θ T γ dt, C 13i = g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] β T Z i γ Θ β 34

35 Z i EZ i β T Z i γ ) } γ β T X i t) Eβ T X i t) β T Z i γ }] g Y i, β T Z i γ ) Eg Y i, β T Z i γ ) β T Z i γ } ] E Θ β T β Z i γ Z i EZ i β T Z i γ ) } ) γ β T X i t) Eβ T X i t) β T Z i γ }], with its lth element C 13l t) satisfies EC 13li t)} =, EC 13li t)} <, C 13li ) 2 <, a.s. y Condition A6). Therefore, using the similar argument as those led to S34) and Lemma 3 we have G 11 G 12 = O p h n 1 logn)} = o p h ). Further, G 12 G 13 = O p h q ) y Condition A5). As a result, we have G 12 G 13 = o p G 11 ) = o p h ) S35) and G 13 2 N G 13 = NO p G 11 ) = O p h 1/2 ) S36) y S34) and we can write G 1 = G T 13n 1/2 γ β ) γ } + o p 1). Comining with the result that G 2 = O p h q ) and the fact that G = G + G 1 + G 2, we have G = G + G T 13n 1/2 γ β ) γ } + o p 1) = O p 1). Further, G 1 = O p h 2 ), G 2 = O p h 2 ) and G 3 = o p 1), G = G +G 1 +G 2 +G 3, so we have G = G + G T 13n 1/2 γ β ) γ } + o p 1) = O p 1). S37) 35

36 The Asymptotic Property for H vector, and Note that H is a finite dimensional B r ) γβ ) α ) B r ) γβ ) B r )γ + B r )γ α ) = O p n 1/2 h 1/2 + h q ) = o p 1). Now y the fact that the kernel estimators are uniformly consistent, we have H = H + H 1 + o p 1), where H = E g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ) Θ β α X i ) Eα X i ) β T α X i )} ] / β T }Θ T β, S38) H 1 = n 1 g Y i, β T Z i γβ )} Eg Y i, β T Z i γβ )} β T Z i γβ )] ) Θ β Z i EZ i β T Z i γβ )} ] γβ )}/ γ T β )] γβ)/ β T β=β = G T 11 γ β)/ β T β=β 1 + o p 1)} = G T 13 γ β)/ β T β=β 1 + o p 1)} = G T 13P1 + o p 1)} = G T 13P + o p 1), where P is defined in S53). S39) The second equality holds ecause γβ ) is consistent to γ. The third equality holds y S35). The fourth equality holds y S52). The last equality holds ecause G 13 2 = O p h 1/2 ) y S36), and P 2 = O p h 1/2 ) y S54). Comining S37), S38), S39) and S32), we have n β β ) S4) 36

37 = H 1 G = H + H 1 + o p 1)} 1 G + G T 13n 1/2 γ β ) γ } + o p 1)} = H + G T 13P + o p 1)} 1 G + G T 13L + o p 1)} = O p 1) where L is defined in Theorem 1. Part II Now note that we can write P = P 1,..., P J 1 ), where P l is the lth column of P, l = 1,..., J 1, which can e written as. P l = E X argmin Pl il t)α t)dt + B r t) T Θ T P γ l β T X i t)dt Hence, } ]) 2 E B r t) T Θ T P γ l β T X i t)dt β T α X i ) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]/ β T α X i )} }. B r ) T Θ T γ P l = argmin Br ) T Θ T γ P l E X il t)α t)dt + B r t) T Θ T P γ l β T X i t)dt } ]) 2 E B r t) T Θ T P γ l β T X i t)dt β T α X i ) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]/ β T α X i )} }. Now let δ l C q, 1], δ l ) = 1 and δ l ) 2 = E X argmin δl il t)α t)dt + β T δ l t)x ic t)dt]) S41) } gc Y i, β T α X i )} where recalling that gc Y i, β T α X i )} is defined efore Theorem 2, and X ic s) = X i s) EX i s) β T α X i )}. We further define α c X i ) = α X i ) 37

38 Eα X i ) β T α X i )}. As the numer of spline knots goes to infinity, B r ) T Θ T γ P l δ l ) = O p h q ). Hence GT 13P l E gc Y i, β T α X i )}Θ β α c X i ) β T X ic t)δ l t)dt] = o p 1). S42) Further, the Gâteaux derivative of the target function in S41) at δ l = δ l in the direction w t) satisfies = E X il t)α t)dt + gc Y i, β T α X i )}/ v v= = 2E X il t)α t)dt gc Y i, β T α X i )} ]) 2 β T X ic t)δ l t) + v T wt)}dt ] β T X ic t)wt)dt +2 δ l t)θ β Eα c X i ) g Y i, β T α X i )}X T ict)]β dt ] = 2E X icl t)α t)dt gc Y i, β T α X i )} β T X ic t)wt)dt +2 δ l t)θ β Eα c X i ) g Y i, β T α X i )}X T ict)]β dt, where w t) is defined in 7), X icl is the lth element of X ic. The second equality holds y the definition of w t) in 7). The last equality holds ecause =. E EX il t) β T α }α t)dt gc Y i, β T α X i )} This implies E Θ β α c X i ) gc Y i, β T α X i )} = E X icl t)α t)dt gc Y i, β T α X i )} ] β T X ic t)δ l t)dt ] β T X ic t)wt)dt ] β T X ic t)wt)dt. This holds for each l = 1,..., J 1. Comine this result with S42), we have GT 13P + E gc Y i, β T α X i )} β T X ic t)wt)dtα c X i ) T Θβ] T

39 = o p 1). Next, y S29) and S3), where ĈQβ ) and C Q β ) in S29) and S3) are defined in S25), we can write where L = argmin Ln 1 gc Y i, β T α X i )} 1 + Further, y aove equality we have Let B r ) T Θ T γ L = argmin Br ) T Θ T γ L n 1 n 1/2 L = L1 + o p 1)}, S43) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] gc Y i, β T α X i )} 1 + ]) 2 B r t) T Θ T Lβ T γ X ic t)dt gc Y i, β T α X i )}. g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] ]) 2 B r t) T Θ T Lβ T γ X ic t)dt gc Y i, β T α X i )}. η ) = argmin η ) n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] gc Y i, β T α X i )} 1 ]) 2 } + ηt)β T X ic t)dt gc Y i, β T α X i ). S44) As the numer of spline knots goes to infinity, B r ) T Θ T γ L ) η ) = O p h q ). Comining with S43) we have GT 13L E Θ β α c X i ) ] gc Y i, β T α X i )}β T X ic t) n 1/2 ηt)dt = o p 1). S45) 39

40 Now taking the Gâteaux derivative of the target function in S44) at η in the direction of wt), we have = n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )]) gc Y i, β T α X i )} 1 + ηt) + v T wt)}β T X ic t)dt ] gc Y i, β T α X i ) / v v= = 2n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )]) ] gc Y i, β T α X i )} 1 + ηt)β T X ic t)dt ) gc Y i, β T α X i )} β T X ic t)wt)dt = 2n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )]) } β T X ic t)wt)dt g c Y i, β T α X i )} = 2n ηt)e β T X ic t)dt ] β T X ic t)wt)dt + o p n 1/2 ) g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )]) } β T X ic t)wt)dt + 2 ηt)θ β Eα c X i ) ) g Y i, β T α X i )}X T ict)]β dt + o p n 1/2 ). The last equality holds y the definition of wt). This implies n 1 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] = +o p n 1/2 ) ) β T X ic t)wt)dt ηt)θ β Eα c X i ) g Y i, β T α X i )}X T ict)]β dt } 2 4

41 Comining with S45), we have GT 13L + n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] = o p 1). β T X ic t)w t)dt) 2 As a result, y S4) the population asymptotic forms of n β β ) can e written as n β β ) = H + G T 13P + o p 1)} 1 G + G T 13L + o p 1)} ] = H E gc Y i, β T α X i )} β T X ic t)wt)dtα c X i ) T Θ T β } 1 +o p 1) G n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )]) = A 1 B + o p 1), } ] β T X ic t)wt)dt + o p 1) where A = E gc Y i, β T α X i )}Θ β α c X i )} 2 ] ] ) E gc Y i, β T α X i )} β T X ic t)wt)dtα c X i ) T Θ T β and B = n 1/2 g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}] } Θ β α c X i ) β T X ic t)wt)dt. If g = g, then Eg Y i, β T α X i )} β T α X i )}] =, E g c Y i, β T α X i )} β T αx i )] = Eg 2 Y i, β T α X i )} β T αx i )] and w t) = w t) as defined in S48). Hence we have n β β) = A 1 B + o p 1) 41

42 where and A = Eg 2 Y i, β T α X i )}Θ β α c X i )} 2 ] ] E g 2 Y i, β T α X i )} β T X ic t)w t)dtα c X i ) T Θ T β B = n 1/2 gy i, β T α X i )}] Θ β α c X i ) = S oeff, } β T X ic t)w t)dt where S oeff is defined in S47). With simple calculation, we can see that A = varb ). Hence, when g is correctly specified, β is the efficient estimator. This proves the result. S3.3 Proof of Theorem 3 Proof: sup B r t) T γ β) α t) t,1] = sup B r t) T Θ T γ γ β) α t) t,1] sup B r t) T Θ T γ γ β) B r t) T Θ T γ γ β ) + sup B r t) T Θ T γ γ β ) B r t) T Θ T γ γ t,1] t,1] + sup B r t) T γ α t) t,1] γ β)/ β T β=β β β ) + γβ ) γ + O p h q ) = O p n 1/2 h 1/2 = O p n 1/2 h 1/2 + h q ) = O p n 1/2 h 1/2 ). + h q ) + γβ ) γ The third line holds y Lemma 1 and Condition A5). The fourth line holds ecause γ β)/ β T β=β β β ) = P 1 + o p P 1 ) S46) 42

43 y Lemma 4, where P 1 = E Θ γ Z i EZ i β α X i }] T β where g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ] ) β T α X i ) )} 1 ) β T Z i EZ i β T α X i )}]Θ T γ M β β ). M = EΘ γ Zi EZ i β T α X i )} ] T β α X i ) T Θ T β g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )}]/ β T α X i )}) is a d γ 1 d β dimensional matrix. Now note that P 1 P 1 2 y the fact that the vector sup norm is less than its L 2 norm, we otain P 1 E Θ γ Z i EZ i β α X i }] T β g Y i, β T α X i )} Eg Y i, β T α X i )} β T α X i )] ] ) β T α X i ) )} 1 ) β T Z i EZ i β T α X i )}]Θ T γ M 2 β β ) 2 = O p h 1 )O p h 1/2 )O p n 1/2 ) = O p n 1/2 h 1/2 ). The second to the last equality holds ecause of S55) and the root n convergence of β. In addition, M 2 = O p h 1/2 ) y S56). The fifth equality in S46) holds ecause y choosing a to e a unit vector with 1 at one entry in Theorem 1, we can otain that each entry of γβ ) γ is of order O p n 1/2 h 1/2 ), and thus γβ ) γ = O p n 1/2 h 1/2 ). The last equality in S46) holds y Condition A4). 2 43

44 S4 Necessary propositions and lemmas Proposition S4.1. The efficient score function of model 2) is S eff = S T effβ, ST effγ )T, where S effβ = f 2Y, β T Zγ) fy, β T Zγ) I J 1, )Z EZ β T Zγ)}γ, S effγ = f 2Y, β T Zγ) fy, β T Zγ), I d γ 1)Z EZ β T Zγ)} T β. Here f 2 stands for the derivative of f with respect to the second variale. Proof: We first calculate the score vector S = S T β, ST γ ) T. Taking derivative of the log likelihood of one oservation with respect to the parameters in β and γ, we otain S β = f 2Y, β T Zγ) fy, β T Zγ) I J 1, )Zγ, S γ = f 2Y, β T Zγ) fy, β T Zγ), I d γ 1)Z T β. We first show that S eff Λ. This is verified y noting that S eff = S ES Y, β T Zγ), and S satisfies ES Z) = = ES β T Zγ). In addition, we also have ES Y, β T Zγ) Λ f Λ since EES Y, β T Zγ) β T Zγ} =. Proposition S4.2. Under model 1), the efficient score function for β is } S oeff = gy, β T αx)} Θ β α c X) β T X c t)w t)dt S47) where w t) is a J 1)-dimensional function that satisfies = Θ β Eα c X)g 2 Y, β T αx)}x T c t)]β Eβ T X c s)g 2 Y, β T αx)}x T c t)β]w s)ds. S48) Here we use the suindex o to indicate quantities calculated under the original model 1). 44

45 Proof: We first make the oservation that Λ is the nuisance tangent space orthogonal complement of the B-spline approximated model 2). Follow the same arguments as those lead to Proposition 3, it can e shown that the spaces corresponding to Λ x, Λ f for model 1) are Λ ox = f Xt)} : f such that Ef) = ], Λ of = f Y, β T αx) } : f such that E f Xt)} = E f β T αx) } = ]. In addition, the nuisance tangent space with respect to αt) is Λ oα = g Y, β T αx) } ] β T Xt)wt)dt : wt) R J 1. After projecting Λ oα to Λ ox and Λ of to get the residual, we otain Λ oα Λ oα Π Λ oα Λ ox + Λ of ) = g Y, β T αx) } ] β T X c t)wt)dt : wt) R J 1. We can easily see that with respect to the model in 1), the nuisance tangent space is Λ = Λ ox Λ of Λ oα. We can see that the score function with respect to the parameters in β is S oβ = gy, β T αx)}θ β αx) = gy, β T αx)} Θ β α c X) +gy, β T αx)} β T X c t)w t)dt +gy, β T αx)}θ β EαX) β T αx)}, } β T X c t)w t)dt where the second and third summands of S oβ elong to Λ oα and Λ of respectively. We can also verify that the first summand of S oβ is simultaneously orthogonal to Λ ox, Λ of and Λ oα, hence the efficient score of model 1) is indeed as given in S47). Lemma 1. There is a constant D r > such that for each spline d γ k=1 c kb rk, and for each 1 p D r c p d γ k=1 c k B rk p c p, 45

46 where c = c k t k t k r )/r} 1/p, k = 1,..., d γ } T. Proof: This is a direct consequence of Theorem on page 145 in DeVore and Lorentz 1993). Lemma 2. Let u e a d γ -dimensional vector with u 2 = 1. There exist positive constants D 1, D 2, D 3, D 4 such that and in proaility. D 1 h u T Cβ)u D 2 h, D2 1 h 1 u T Cβ) 1 u D1 1 h 1, D 3 h u T Ĉβ)u D 4 h, D4 1 h 1 u T Ĉβ) 1 u D3 1 h 1 Proof: First note that y the Cauchy-Schwartz inequality, Lemma 1 and Condition A6), we have E u T = E } 2 ] B r t)β T Xt)dt dγ u k B rk t) k=1 dγ u k B rk t) k=1 u 2 2O1) = Oh ), }2 }2 dt β T Xt)} 2 dt } dte β T Xt)} 2 dt where u = u k t k t k r )/r} 1/2, k = 1,..., d γ } T, whose L 2 norm is of order Oh 1/2 ). Thus we have u T Cβ)u D 2 h for some positive constant D 2 <. As shown in 28) in Cardot et al. 23), since the eigenvalues of the covariance operator Γβ) are strictly positive, there is a positive constant D such that < Γβ)φ, φ > D φ 2, for φ H. 46

47 Note that B T r u H, so we have u T Cβ)u =< Γβ)B T r u, B T r u > D B T r u 2 D 1 u 2 h = D 1 h for a positive constant D 1 y Lemma 1 and Condition A4). Therefore, D 1 h u T Cβ)u D 2 h. And so D 1 2 h 1 Theorem 1.19 in Chatelin 1983), we have Ĉβ) Cβ) 2 sup 1 l d γ u T C 1 β)u D 1 d γ k=1 1 h 1 Γ n Γ 2 < B rk, B rl >. Further, with As shown in Cardot et al. 23), Lemma 5.3 in Cardot et al. 1999) implies Γ n Γ 2 = o p n h 1)/2 ). Further, y the property of B-spline asis, we have when k l > r, B rk B rl =. Therefore, sup 1 l dγ dγ k=1 < B rk, B rl > = Oh ), which implies Ĉβ) Cβ) 2 o p h n h 1)/2 ). S49) Now ecause h < 1, comine with the result that D 1 h u T Cu D 2 h, y the triangular inequality we otain and Thus, D 3 h u T Ĉβ)u = u T T Cβ)u + u Ĉβ) Cβ)}u D 2 h + u 2 Ĉβ) Cβ)}u 2 D 2 h + Ĉβ) Cβ)} 2 u 2 = D 2 h + o p h ) u T Ĉβ)u = u T T Cβ)u + u Ĉβ) Cβ)}u D 1 h u 2 Ĉβ) Cβ)}u 2 D 1 h Ĉβ) Cβ)} 2 u 2 = D 1 h + o p h ). u T Ĉβ)u D 4 h in proaility for some positive constant D 3, D 4 <. And so D 1 4 h u T Ĉβ) 1 u D 1 3 h 1 proves the result. in proaility. This 47

48 The result in Lemma 2 can e generalized to any vector valued random function with almost surely ounded L 2 norm and positive definite second moment operator. Therefore, we have the following corollary. Corollary 1. Let Q i t) e a d q -dimensional vector valued random function with ounded L 2 norm, and let v e an aritrary d q -dimensional vector. Let Γ Q v) e the second moment operator generated y v T Q i ), i.e., Γ Q v)φt) = E< v T Q i, φ > v T Q i t)}. Assume Γ Q v) is strictly positive definite. We define the matrix C Q v) to e a d γ d γ matrix with its k, l) element E B rk t)v T Q i t)dt B rl t)v T Q i t)dt}, and ĈQv) to e a d γ d γ matrix with its k, l) element n 1 B rk t)v T Q i t)dt B rl t)v T Q i t)dt. Let u e a d γ -dimensional vector with u 2 = 1. Then there are positive constants D 5, D 6, D 7, D 8 such that and in proaility. And D 5 h u T C Q v)u D 6 h, D 1 6 h 1 u T C 1 Q v)u D 1 5 h 1, D 7 h u T Ĉ Q v)u D 8 h, D 1 8 h 1 u T Ĉ 1 Q v)u D 1 7 h 1, ĈQv) C Q v) 2 o p h n h 1)/2 ). The proof of Corollary 1 follows the same arguments as those in the proof of Lemma 2 hence is omitted. 48